The general form of univariate cubic equation is
x3+sx2+tx+u=0
If we do an abscissa translation y=x+s/3, then we can eliminate the quadratic term of the equation.
Let's go So we just need to consider the shape.
x3=px+q
Cubic equation of.
Suppose that the solution X of the equation can be written in the form of x=a-b, where A and B are undetermined parameters.
Substitute it into the equation and we have it.
a3-3a2b+3ab2-b3=p(a-b)+q
Complete sorting
a3-b3 =(a-b)(p+3ab)+q
According to the theory of quadratic equation, A and B can be properly selected, so that when x=a-b,
3ab+p=0. So the above formula becomes
a3-b3=q
Multiply both sides by 27a3, and you get
27a6-27a3b3=27qa3
P=-3ab indicates that
27a6 + p = 27qa3
This is a quadratic equation about a3, so A can be solved. Then we can solve the roots of b and x.
In addition to finding the root formula and factorization, we can also use the mirror method to solve the mean value theorem. Many higher-order equations can't get exact solutions. For this kind of equation, we can use dichotomy and tangent method to get approximate solutions with arbitrary accuracy. See Tongji Fourth Edition Advanced Mathematics.
The root formula of the unary cubic equation can't be worked out by ordinary deductive thinking. The standard unary cubic equation of AX 3+BX 2+CX+D+0 can only be formalized into a special type of X 3+PX+Q = 0 by using a matching method similar to the root formula of the unary quadratic equation.
The solution of the solution formula of the univariate cubic equation can only be obtained by inductive thinking, that is, the form of the root formula of the univariate cubic equation is summarized according to the form of the root formula of the univariate quadratic equation and the special higher order equation. The formula for finding the root of a univariate cubic equation in the form of x 3+px+q = 0 should be in the form of X = A (1/3)+B (1/3), which is the sum of two square roots. Summarized the form of the root formula of the univariate cubic equation, and the next step is to find the content of the square, that is, to represent a and b by P and Q, as follows:
(1) can get the simultaneous cube of X = A (1/3)+B (1/3).
(2)x^3=(a+b)+3(ab)^( 1/3)(a^( 1/3)+b^( 1/3))
(3) Because X = A (1/3)+B (1/3), (2) can be changed to
X 3 = (a+b)+3 (ab) (1/3) x, transpositions are available.
(4) x 3-3 (ab) (1/3) x-(a+b) = 0. Comparing the univariate cubic equation with the special type x 3+px+q = 0,
(5)-3 (AB) (1/3) = P, -(A+B) = Q, simplified.
(6)A+B=-q,AB=-(p/3)^3
(7) In this way, the roots of the univariate cubic equation are formulated into the roots of the univariate quadratic equation, because A and B can be regarded as the two roots of the univariate quadratic equation, and (6) is Vieta's theorem about the two roots of the univariate quadratic equation in the form of ay 2+by+c = 0, namely.
(8)y 1+y2=-(b/a)，y 1*y2=c/a
(9) Comparing (6) and (8), we can make A = Y 1, B = Y2, Q = B/A,-(p/3) 3 = c/a.
(10) because the formula for finding the root of the unary quadratic equation of type ay 2+by+c = 0 is
y 1=-(b+(b^2-4ac)^( 1/2))/(2a)
y2=-(b-(b^2-4ac)^( 1/2))/(2a)
Can become
( 1 1)y 1=-(b/2a)-((b/2a)^2-(c/a))^( 1/2)
y2=-(b/2a)+((b/2a)^2-(c/a))^( 1/2)
Substitute a = y 1, b = y2, q = b/a, -(p/3) 3 = c/a in (9) into (1 1).
( 12)a=-(q/2)-((q/2)^2+(p/3)^3)^( 1/2)
b=-(q/2)+((q/2)^2+(p/3)^3)^( 1/2)
(13) substitute a and b into X = A (1/3)+B (1/3).
( 14)x=(-(q/2)-((q/2)^2+(p/3)^3)^( 1/2))^( 1/3)+(-(q/2)+((q/2)^2+(p/3)^3)^( 1/2))^( 1/3)
Postscript:
1.( 14) is only the real root solution of the univariate cubic equation. According to the cubic equation of Vieta's theorem, there should be three roots. But according to the cubic equation of Vieta's theorem, only one root is needed, and the other two roots are easy to find. Because the calculation is too complicated and this problem has been solved in history, I don't want to spend too much energy on it. I just want to test my intelligence, so as long as I solve the other two roots, I won't spend any effort to solve them.
Secondly, I have solved the univariate quartic equation in a similar way, that is, assuming that the root of the univariate quartic equation is in the form of x = a (1/4)+b (1/4)+c (1/4). I seem to have solved it once, but then. However, I think if we can further summarize the forms of A, B and C, we should be able to find the root formula of the quartic equation with one variable. Because the calculation is too complicated and this problem has been solved by the ancients, I have never been able to finish this work.
3. I gained an experience by solving the root formula of cubic equation with one variable. For problems that can't be solved by deduction (that is, direct reasoning), we can often get good results by changing our thinking and inducing (inductive analogy of solutions to simple and special similar problems). In fact, human beings often solve problems in this way, and great scientists become great scientists in this way. The quadratic formula ax 2+bx+c = 0.
X=[-b+ radical symbol (b-4ac)]/2a
X=[-b- radical sign (b 2-4ac)]/2a
When b 2-4ac > 0 point.
This equation has two unequal roots.
When b 2-4ac = 0
This equation has roots.
When b 2-4ac
The equation has no real number solution. Binary quadratic equation has no formula, so we can only adopt different methods according to different types of questions:
The first category: a system of equations consisting of binary linear equations and binary quadratic equations,
a 1x+b 1y+c 1 = 0( 1)
a2x^2+b2xy+c2y^2+d2x+e2y+f2=0(2)
By substituting elimination, it can be transformed into a quadratic equation of one variable, which generally has two sets of solutions.
The second type: an equation group consisting of two binary quadratic equations.
a 1x^2+b 1xy+c 1y^2+d 1x+e 1y+f 1=0
a2x^2+b2xy+c2y^2+d2x+e2y+f2=0
(1) If the left side of a binary quadratic equation can be factorized, then this equation will factorize it into two binary linear equations, which will be combined with another set of equations to form two first-class equations, and then substituted into elimination. Generally speaking, this form of equation has four sets of solutions.
(2) If it is an equation group consisting of a quadratic equation with one variable and a quadratic equation with two variables, we can first solve the quadratic equation with one variable and then substitute it into another equation to solve it. Generally, this kind of equations has four groups of solutions.
(3) If a1:a2 = b1:b2 = c1:c2, the quadratic term can be eliminated and the first kind can be solved.
(4) If a1:a2 = b1:b2 = d1:d2 or b1:b2 = c1:c2 = e1:e2, the elimination method can be used to solve the first.
The solution is
=√ 1/3969
= 1/63
still
1/√3969?
Is the denominator in the numerator? Or driving together? Or what? 880